Monday, Oct. 11, 2010 PHYS 1441-002, Fall 2010 Dr. Jaehoon Yu

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PHYS 1441 – Section 002Lecture #10

Monday, Oct. 11, 2010Dr. Jaehoon Yu

• Force of Friction• Application of Newton’s Laws• Example for applications of Newton’s Laws

– Example for Motion without friction– Example for Motion with friction

• Uniform Circular Motion

Today’s homework is homework #6, due 10pm, Tuesday, Oct. 19!!

Announcements• Quiz #2 Results

– Class average: 17/30• Equivalent to: 57/100

– Top score: 30/30• Evaluation policy

– Homework: 30%– Comprehensive final exam: 25%– One better of the two non-comprehensive term exam: 20%– Lab: 15%– Quiz:10%– Extra credit: 10%

• Colloquium this weekMonday, Oct. 11, 2010 PHYS 1441-002, Fall 2010 Dr. Jaehoon

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Monday, Oct. 11, 2010 3PHYS 1441-002, Fall 2010 Dr. Jaehoon Yu

Monday, Oct. 11, 2010 PHYS 1441-002, Fall 2010 Dr. Jaehoon Yu

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Reminder: Special Project for Extra CreditA large man and a small boy stand facing each other on frictionless ice. They put their hands together and push against each other so that they move apart. a) Who moves away with the higher speed and by how much? b) Who moves farther in the same elapsed time?

• Derive fomulae for the two problems above in much more detail than what is shown in this lecture note.

• Be sure to clearly define each variables used in your derivation.

• Each problem is 10 points each.• Due is this Wednesday, Oct. 13.

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Applications of Newton’s Laws

M

Suppose you are pulling a box on frictionless ice, using a rope.

T

What are the forces being exerted on the box?Gravitational force: FgNormal force: nTension force: T

n= -Fg

TFree-body diagram

Fg=MgTotal force: F=Fg+n+T=T

xF

If T is a constant force, ax, is constant

xfv

yF

M

Tax

0ya

x

n= -Fg

Fg=Mg

T

T Max

−Fg + n = yMa 0

vxi + axt=vxi +TM

⎛⎝⎜

⎞⎠⎟t

x f −xi =vxit +12

TM

⎛⎝⎜

⎞⎠⎟t2

What happened to the motion in y-direction?

PHYS 1441-002, Fall 2010 Dr. Jaehoon Yu

Monday, Oct. 11, 2010 PHYS 1441-002, Fall 2010 Dr. Jaehoon Yu

When an object is in contact with a surface there is a force acting on that object. The component of this force that is parallel to the surface is called the friction force.

Friction Force

This resistive force is exerted on a moving object due to

Always opposite to the movement!!

viscosity or other types of frictional property of the medium in or surface on which the object moves.

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Monday, Oct. 11, 2010 PHYS 1441-002, Fall 2010 Dr. Jaehoon Yu

When the two surfaces are not sliding across one anotherthe friction is called static friction.

Static Friction

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The resistive The resistive force exertedforce exertedon the object up to the time just before on the object up to the time just before

the object starts moving.the object starts moving.

Monday, Oct. 11, 2010 PHYS 1441-002, Fall 2010 Dr. Jaehoon Yu

The magnitude of the static friction force can have any value from zero up to the maximum value.

MAXs sf f

MAXsf

0 1s is called the coefficient of static friction.

Magnitude of the Static Friction

Kinetic friction applies during the move!!

s NF

What is the unit?

Once the object starts moving, there is NO MORE static friction!!

None

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Monday, Oct. 11, 2010 PHYS 1441-002, Fall 2010 Dr. Jaehoon Yu

Note that the magnitude of the frictional force does not depend on the contact area of the surfaces.

MAXsf s NF

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Monday, Oct. 11, 2010 PHYS 1441-002, Fall 2010 Dr. Jaehoon Yu

Static friction opposes the impending relative motion betweentwo objects.

kf 0 1k is called the coefficient of kinetic friction.

Kinetic Friction

Kinetic friction opposes the relative sliding motions that is happening. The resistive force exerted on the object during its movement.

k NF

What is the direction of friction forces? opposite to the movement

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Normally much smaller than static friction!!

Monday, Oct. 11, 2010 PHYS 1441-002, Fall 2010 Dr. Jaehoon Yu

Coefficient of Friction

What are these?

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Monday, Oct. 11, 2010

Forces of Friction Summary

furs ≤s F

urN

Resistive force exerted on a moving object due to viscosity or other types frictional property of the medium in or surface on which the object moves.

Force of static friction, fs:

Force of kinetic friction, fk

The resistive force exerted on the object until just before the beginning of its movement

The resistive force exerted on the object during its movementNkkf F

ur ur

These forces are either proportional to the velocity or the normal force.

Empirical Formula

What does this formula tell you?

Frictional force increases till it reaches the limit!!

Beyond the limit, the object moves, and there is NO MORE static friction but kinetic friction takes it over.

Which direction does kinetic friction apply? Opposite to the motion!PHYS 1441-002, Fall 2010 Dr. Jaehoon

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Monday, Oct. 11, 2010 13

Look at this problem again…

M

Suppose you are pulling a box on a rough surfice, using a rope.

T

What are the forces being exerted on the box?Gravitational force: FgNormal force: nTension force: T

n= -Fg

TFree-body diagram

Fg=MgTotal force: F=Fg+n+T+Ff

xF

If T is a constant force, ax, is constant

xfv

yF

ax T −Ff

M

0ya

x

n= -Fg

Fg=Mg

T

T −Ff Max

−Fg + n = yMa 0

vxi + axt=vxi +T −Ff

M⎛⎝⎜

⎞⎠⎟t

x f −xi = vxit +12

T −Ff

M⎛⎝⎜

⎞⎠⎟t2

PHYS 1441-002, Fall 2010 Dr. Jaehoon Yu

Friction force: Ff

Ff

Ff

n Fg

Monday, Oct. 11, 2010 PHYS 1441-002, Fall 2010 Dr. Jaehoon Yu

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Example of Using Newton’s LawsA traffic light weighing 125 N hangs from the cable tied to two other cables fastened to a support. The upper cables make angles of 37.0o and 53.0o with the horizontal plane. Find the magnitudes of the tensions in the three cables.

F ur

Free-bodyDiagram

53o37o

x

y

T1

37o

T2

53o

T3

−T1 cos 37o( ) + T2 cos 53o( ) = 0 1T

T2 sin 53o( ) + 0.754 ×sin 37o( )⎡⎣ ⎤⎦=1.25T2 =125N

T2=100N;

03

1

i

iixx TF

03

1

i

iiyy TF

Newton’s 2nd law

x-comp. of net force

y-comp. of net force

0

2

cos 53

cos 37T

o

o 20.754 T

T1=0.754T2 =75.4N

Tur

1+Tur

2 +Tur

3 mar0

Monday, Oct. 11, 2010 PHYS 1441-002, Fall 2010 Dr. Jaehoon Yu

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Example w/o FrictionA crate of mass M is placed on a frictionless inclined plane of angle θ. a) Determine the acceleration of the crate after it is released.

Fur=

Free-bodyDiagramθ

x

yM

da

Fg

nn

F= -MgθSupposed the crate was released at the top of the incline, and the length of the incline is d. How long does it take for the crate to reach the bottom and what is its speed at the bottom?

d =

yF

xF

ax =gsinθx

y

t =2d

gsinθ

xfv

x

y

Fur

g +nr= am

xMa gxF θsinMg

May = gyFn n −mgcosθ = 0

vixt +12

axt2 =

1

2gsinθ t2

vix + axt=gsinθ2d

gsinθ 2dgsinθ

vxf = 2dgsinθ

Monday, Oct. 11, 2010

Example w/ FrictionSuppose a block is placed on a rough surface inclined relative to the horizontal. The inclination angle is increased till the block starts to move. Show that by measuring this critical angle, θc, one can determine coefficient of static friction, μs.

F ur

Free-bodyDiagram

θx

yM a

Fg

nn

F= -Mgθ

Fs=μsn

yF

xF

s

Net force

x comp.

y comp.

sf

n

x

y

M a r

Fur

g + nr+ f

urs

Fgx − fs sfMg θsin 0 ns cMg θsin

yMa gyFn cMgn θcos 0 gyF cMg θcos

n

Mg cθsin c

c

Mg

Mg

θθ

cos

sincθtan

PHYS 1441-002, Fall 2010 Dr. Jaehoon Yu

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